《ACM程序设计》书中题目P

Description
FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1’s. All integers are not greater than 1000.

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output

13.333
31.500

价值/代价，排序，贪心，可怜的翁恺老师。。。。

#include<bits/stdc++.h>using namespace std;struct Trade{    int j,f;    double percent;}mouse[3001];bool cmp(Trade a,Trade b){    return a.percent>b.percent;}int main(){   int m,n,i,sum,ans,temp;   double cnt;   while(~scanf("%d%d",&m,&n))   {         sum=0;ans=0;temp=0;       if(m==-1&&n==-1)           break;       for(i=0;i<n;i++)       {           scanf("%d%d",&mouse[i].j,&mouse[i].f);           mouse[i].percent=(double)mouse[i].j/mouse[i].f;       }       sort(mouse,mouse+n,cmp);       for(i=0;i<n;i++)       {           sum+=mouse[i].f;           if(sum<=m)           {               ans+=mouse[i].j;               temp+=mouse[i].f;            }           else           {               sum=m-temp;               break;           }       }       cnt=(double)mouse[i].j*((double)sum/mouse[i].f);         printf("%.3lf\n",cnt+ans);   }   return  0;}